A question about the quotient of a $K$-algebra by its radical. Let $A$ be a $K$-algebra and $B=A/\operatorname{rad} A$, where $\operatorname{rad}A$ is the radical of $A$ (intersection of all maximal right ideals of $A$).  Let $e$ be an idempotent of  $A$ and $\bar{e}=e+\operatorname{rad} A$. How to show that $eA/\operatorname{rad} eA$ is isomorphic to $\bar{e}B$? We have $\bar{e}B=(e+\operatorname{rad} A)(A/\operatorname{rad} A)$. The elements in $\bar{e}B$ is of the form $ea+\operatorname{rad} A$, where $a \in A$. But the elements of $eA/\operatorname{rad} eA$ is of the form $ea+\operatorname{rad} eA$, where $a \in A$. This question comes from the reading of the book (page 21, line 1 of the proof of Proposition 4.5 of the book Elements of the Representation Theory of Associative Algebras: Volume 1). Another question is how to show that $\operatorname{rad} eA = eA \operatorname{rad} A = e \operatorname{rad} A$? Thank you very much.



 A: Prequel: How to show that $\operatorname{rad}eA=eA\operatorname{rad} A=e\operatorname{rad}A\subseteq \operatorname{rad} A$?
The last two (in)equalities follow from the fact that $\operatorname{rad}A$ is a left ideal. For the first equality you just note that $\operatorname{rad} A=\operatorname{rad} eA\oplus \operatorname{rad} (1-e)A$ (in fact it holds that $\operatorname{rad}$ is compatible with every direct sum decomposition) Furthermore note $A\operatorname{rad} A=\operatorname{rad} A$ and that $A\operatorname{rad} A=eA\operatorname{rad} A\oplus (1-e)A\operatorname{rad} A$ (again in fact this compatibility holds for every direct sum decomposition). Now if you compose all this equalities and use that this is a direct sum decomposition you get the result. (More details in the case of left modules can e.g. be found in Ringel, Schröer: Representation theory of algebras I.
Original Answer: I assume (now) you know the following: $\operatorname{rad} eA=e\operatorname{rad} A\subseteq \operatorname{rad} A$.
Then define just the obvious maps between $e(A/\operatorname{rad} A)$ and $eA/\operatorname{rad} eA$.
If just have to show well-definedness (that they are mutually inverse and $A$-module homomorphisms is quite obvious.)
So suppose $ea+x=eb+y$ with $a,b\in A$, $x,y\in \operatorname{rad} A$. Then $e(ea+x)=e(eb+y)$, hence $ea+ex=eb+ey$ and thus $ea=eb \mod \operatorname{rad} eA$.
For the other direction suppose $ea+ex=eb+ey$. Then since $ex, ey\in \operatorname{rad} A$ also the inverse map is well-defined.
